Katie arithmetic mean

Let the numbers on Katie's list be $x_1,\cdots,x_n$ . We wish to maximise $\overline{x}=\frac{1}{n} \sum x_i$ subject to the condition $\sum x_i^2 - x_i=0$ .

Solution 1: Lagrange multipliers

For a fixed positive integer $n$ , let $F(x_1,\cdots,x_n,L) = \frac{1}{n} \sum x_i + \frac{L}{n}\sum x_i^2 - x_i$

The maximum of $\overline{x}$ subject to the constraint is achieved when each of the partial derivatives of $F$ is zero. For each $i$ , we have $\frac{\partial F}{\partial x_i} = \frac{1}{n} + \frac{L}{n}(2x_i-1)=0$

The trick here is to note that this means (for an extremum) all of the $x_i$ must be equal - in fact, $x_i=\overline{x}$ for each $i$ . Hence $\overline{x}^2-\overline{x}=0$ , so $\overline{x}=0$ or $\overline{x}=\boxed{1}$ .

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Solution 2: Geometry

The condition can be rewritten as $\sum \left(x_i - \frac12 \right)^2=\frac{n}{4}$ . This is the equation of an $n$ dimensional sphere, $S_c$ centred at $(\frac12,\frac12,\cdots,\frac12)$ with radius $\sqrt{\frac{n}{4}}$ .

Due to the condition, maximising the arithmetic mean of the values is the same as maximising $\sum x_i^2=r^2$ . This is also an $n$ dimensional sphere, $S$ , centred at $O$ .

The point $(x_1,\cdots,x_n)$ lies on both spheres. In order to maximise $r$ , the spheres must be tangent to each other ( $S_c$ is inside $S$ ). The point of tangency must lie on the line joining their centres; hence again we find all the $x_i$ are equal, and the rest of the derivation is as in solution 1.

The list may contain the following numbers : $(0,0,0,....),(0,1),(0,0,1),...(0,0,0,....,1),(1,1),(0,1,1),....$ . Of these, the list $(1,1)$ has the largest mean, which is $\boxed 1$ .